A quasitopos is a finitely cocomplete locally cartesian closed category with a terminal object (hence finite limits) and a classifier for strong subobjects.
A monomorphism $m:A \to B$ is called strong if for any epimorphism $e:C \to D$ and any two morphisms $f:C \to A$ and $g:D \to B$ for which $m \circ f=g \circ e$, there is a (necessarily unique) morphism $h:D \to A$ for which $m \circ h=g$ and $h \circ e=f$.
A classifier for strong subobjects, then, is a morphism $\top:1 \to \Omega$ for which given any strong monomorphism $m:A \to B$, there is a unique "characteristic" or "indicator" morphism $\chi_m:B \to \Omega$ for which $m$ is a pullback of $\top$ along $\chi_m$. Equivalently, $m$ is an equalizer of $\chi_m$ and $\top \circ !:B \to 1 \to \Omega$, so every strong monomorphism is regular (and conversely, regular monomorphisms are always strong).
A classifier for regular subobjects could be defined similarly. So, if one defines a "strong quasitopos" and a "regular quasitopos" to mean a finitely cocomplete locally cartesian closed category with a terminal object (hence finite limits) and a classifier for strong and regular subobjects respectively, is every "regular quasitopos" in fact a "strong quasitopos"?
If not, then a counterexample must contain a non-regular strong monomorphism that obviously cannot have a "characteristic" or "indicator" morphism.
At Is there a topos-like category that classifies regular subobjects?, there is no clear proof on why replacing "strong" with "regular" in the definition of "quasitopos" would or would not make a difference.
The answer is no for the question in the title, yes for the question in the body: any regular quasitopos is a strong quasitopos. This is because if a class of monomorphisms containing all sections (i.e. all split monomorphisms) has a classifier along which its retraction has a pushforward, then it is a class of regular monomorphisms closed under composition (see below for details). In particular, regular monomorphisms are closed under composition in a locally cartesian closed category with a regular subobject classifier. Proposition 12.5 Oswald Wyler's Lectures on Topoi and Quasitopoi then implies any strong monomorphism with a cokernel pair is regular. Thus any regular quasitopos is a strong quasitopos
Suppose a class of monomorphisms $\mathcal M$ has a classifier $\tilde1\leftarrowtail1$ in the sense that up to equivalence each monomorphism $X\leftarrowtail U\in\mathcal M$ is the pullback of $\tilde1\leftarrowtail1$ along a unique morphism $X\to\tilde1$. Note that in that case $1$ is subterminal, i.e. for each $X$ there is at most one morphism $X\to1$, because $\mathrm{id}_X\colon X\to X$ is the pullback along $X\to1\rightarrowtail\tilde1$ of $\tilde1\leftarrowtail 1$ for any $X\to1$.
The pushforward of $X\to1$ along $\tilde1\leftarrowtail1$ is by definition a partial morphism $\tilde X\leftarrowtail X'\to X$ with $\tilde X\leftarrowtail X'\in\mathcal M$ of which any partial morphism $W\leftarrowtail U\to X$ with $W\leftarrowtail U\in\mathcal M$ is up to equivalence the pullback along a unique morphism $W\to\tilde X$. But in fact $X'\to X$ is an isomorphism, so that we have an $\mathcal M$-partial morphism classifier $\tilde X\leftarrowtail X$.
Indeed, the existence of $X\to1$ and $\tilde1\leftarrowtail1$ implies $\mathrm{id}_X\colon X\to X\in\mathcal M$, there must be a unique morphism $X\to\tilde X$ that factors through $\tilde X\leftarrowtail X'$ via a section of $X'\to X$. But then the pullbacks of $\tilde X\leftarrowtail X'\to X$ along $X'\rightarrowtail X$ and $X'\to X'\rightarrowtail X$ for the associated idempotent $X'\to X'$ are both the partial morphism $X'\leftarrow X'\to X$ with $X'\leftarrow X'$ the identity morphism. Therefore the associated idempotent is the identity and $X'\to X$ is an isomorphism. (This is basically the argument of 18.5 and 19.2 of Oswald Wyler's Lectures on Topoi and Quasitopoi without the conditions on $\mathcal M$ imposed in 12.3 and 13.1).
Let $\Omega=\tilde1$ and assume that $\tilde\Omega$ exists. Then a composable pair of morphisms $Z\leftarrowtail Y, Y\leftarrowtail X\in\mathcal M$ arise as the successive pullbacks along a unique morphism $Z\to\tilde\Omega$ of the composable pair $\tilde\Omega\leftarrowtail\Omega=\tilde1,\Omega=\tilde1\leftarrowtail1$. Consequently, $\mathcal M$ is closed under composition if and only if the composite $\tilde\Omega\leftarrowtail\Omega=\tilde1\leftarrowtail1$ is in $\mathcal M$.
Now because $1$ is subterminal, any morphism $1\to X$ is a section of the (necessarily unique) morphism $X\to 1$. In particular, $\tilde1\leftarrowtail1$ is a section and thus a regular monomorphism if $\mathrm{id}_{\tilde 1}\colon\tilde1\to\tilde1\in\mathcal M$, in which case $\mathcal M$ consists of regular monomorphism. Similarly, the composite $\tilde\Omega\leftarrowtail\Omega=\tilde1\leftarrowtail1$ is a section if $\mathrm{id}_{\tilde\Omega}\colon\tilde\Omega\to\tilde\Omega\in\mathcal M$.
Since identity morphisms are also sections, it follows that if a class $\mathcal M$ contains all sections and has a classifier $\Omega=\tilde1\leftarrowtail1$ along which exists a pushforward of its retraction $\Omega=\tilde1\to1$, then it is a class of regular monomorphisms that is closed under composition.